Answer:
Option A
Explanation:
We have, $f:[a,b]\rightarrow \left[-\frac{\sqrt{3}}{4},\frac{1}{2}\right]$
$f(x)=\begin{bmatrix}1 & 1&1 \\1 & 1+\sin_{}x&1\\1+\cos x&1&1 \end{bmatrix}$
$f(x)=\begin{bmatrix}1 & 1&0 \\0 & \sin_{}x&0\\1+\cos x&-\cos x&-\cos x \end{bmatrix}$
$[C_{3}\rightarrow C_{3}-C_{1},C_{2}\rightarrow C_{2}-C_{1}]$
f(x) =-sin x cos x
$f(x)=-\frac{1}{2} \sin 2x$
$f(x)\in \left[\frac{\sqrt{3}}{4},\frac{1}{2}\right]$
$\therefore$ $-\frac{\sqrt{3}}{4}\leq \frac{-1}{2} \sin 2x \leq \frac{1}{2}$
$\Rightarrow$ $-1\leq \sin 2x \leq \frac{\sqrt{3}}{2}$
$\Rightarrow$ $-\frac{\pi}{4}\leq x \leq \frac{\pi}{6}$
$\therefore$ $a=\frac{-\pi}{4},b=\frac{\pi}{6}$
